Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.

将k个有序链表合并成一个有序链表。

思路，之前做过两个有序链表的合并操作，对于k个链表，我们最先想到的就是能不能转化为我们熟悉的两个链表的合并，可以我们先将k个链表分为两部分，先对这两部分完成链表的有序合并操作，再对合并得到的两个链表进行合并，这是一个递归性质的描述，采用递归很容易实现这个程序，和数组的归并排序很类似。

Accepted Solution：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
	ListNode *mergeKLists(vector<ListNode *> &lists)
	{
		return mergeSort(lists, 0, lists.size() - 1);
	}
	ListNode *mergeSort(vector<ListNode *> &lists, int begin, int end)
	{
		if (begin>end)
			return NULL;
		if (begin == end)
			return lists[begin];

		int mid = (begin + end) / 2;
		ListNode *head1 = mergeSort(lists, begin, mid);
		ListNode *head2 = mergeSort(lists, mid + 1, end);
		ListNode *dummyHead = new ListNode(0),*p = dummyHead;
		while (head1 != NULL&&head2 != NULL)
		{
			if (head1->val<head2->val)
			{
				p->next = head1;
				head1 = head1->next;
				p = p->next;
			}
			else
			{
				p->next = head2;
				head2 = head2->next;
				p = p->next;
			}
		}
		if (head1 != NULL)
			p->next = head1;
		if (head2 != NULL)
			p->next = head2;
		p = dummyHead->next;
		delete dummyHead;
		return p;

	}
};